PROBLEM 17 & 18 April 17 2010

PROBLEM 17 & 18 April 17 2010 - so again R I B 2 =...

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SOLVING PROBLEMS 17 and 18 April 17, 2010 Problem 17 (like 24.01): What is the magnetic field at the edge of a 1 m diameter capacitor having a 1 mm gap charging with a 1 A instantaneous current? What do you know? The charge builds up on the plates at a rate of 1 C/s. From Gauss’ law, the electric field across the capacitor satisfies 0 2 Q R E . The magnetic field according to the Maxwell-Ampere law is due to the displacement current I R E dt d R B 0 2 0 0 2   and thus R I B 2 0 = 4 10 -7 1 A/1m = 0.4 T a) 0.4 kT; b) 0.4 T; c) 0.4 mT; d) 0.4 nT; e) 0.4 T. What is the magnetic field 0.5 m from a long wire carrying 1 A of current? What do you know? The magnetic field is now due to the conduction current in the Maxwell-Ampere law: I R B 0 2
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Unformatted text preview: so again R I B 2 = 4 10-7 1 A/1m = 0.4 T a) 0.4 kT; b) 0.4 T; c) 0.4 mT; d) 0.4 nT; e) 0.4 T. Problem 18 (like 24.08): Radio waves below 10 kHz are apparently not regulated by the FCC. What is the wavelength of a 10 kHz radio wave? What do you know? Lambda=v/f = 299792458 ms-1 /10,000 s-1 = 29979 m = 30 km a) 30 km; b) 30,000 km; c) 3 m; d) 3 km; e) 3 cm. What frequency will one detect when driving at 30 m/s towards a 10 kHz radio source? a) 10,000.001 Hz; b) 10,001.000 Hz; c) 10,000.010 Hz; d) 10,100.100 Hz; e) 10,010.00001 Hz. What do you know? Going towards the source the frequency will be higher by a fractional amount v/c so f = (1+v/c) f = 10000.001 Hz...
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